# Modelling Probabilities on a Game of Tennis

Modelling Probabilities on games of tennis

Introduction:

In this portfolio I shall investigate the different models and probabilities based on the probabilities in the game of tennis. First I will start with the Part 1 of the portfolio where I will be concluding with the expected value and the standard distribution from my results.

I will then take a look at the Non Extended play games where the highest of 7 points can be played. This is will be done with the use of binomial distribution. Then I will calculate the odds of Adam winning the game of tennis and will generalize my model so that I can apply to any other player. After making this model, I will take a look at the extended games where in theory the game could go on forever. This is the stage where I find a model to find the odds of Adam winning the extended games and then will generalize this also.

In the Part 3, I will also test the model for different values of point winning probabilities and find out the odds for each of them and then will then look for patterns from the values of odds that I find.

Finally, I will evaluate the benefits and limitations of the models I come up with.

Part 1:

According to the question, Adam wins twice as many points as Ben does in the game of tennis; therefore, the ratio of points won by Adam to Ben is 2:1 respectively. This shows that Adam wins 23of the points and Ben wins 13of the points.

The distribution of X, i.e., the number of points won be Adam is derived using the binomial distribution, by substituting the variables and the constants, we are able to arrive at the appropriate model for distribution of X, the number of points won by Adam.

We can begin by stimulating a 10-point game. Since distribution of X denotes the number of points scored by A, it can be represented as:

X ~ B (n, p) =>X ~ B (10,23)

Hence according to the formula of Binomial distribution,

P (X=x) = nCx×px× qn-x

* 10x×23x×13n-x

The probability of Adam winning ‘X’ number of points can be found by substituting ‘X’ with the number of points that Adam wins and after solving, the answer will be Adams probability of winning the given number of points that were substituted into the equation. The table below shows the values of the probabilities:

x| P (X=x)[Fraction form]| P (X=x)[Decimal form]|

0| 159049| 0.000017|

1| 2059049| 0.000339|

2| 18059049| 0.003048|

3| 96059049| 0.016258|

4| 336059049| 0.056902|

5| 806459049| 0.136565|

6| 1344059049| 0.227608|

7| 1536059049| 0.260123|

8| 1152059049| 0.195092|

9| 512059049| 0.086708|

10| 102459049| 0.017342|

To check whether the fractions found are correct, they all are added up, as their sum should be 1. Therefore here I will add all the above fractions:

(159049+ 2059049+ 18059049+ 96059049+ 336059049+ 806459049+ 1344059049+ 1536059049+ 1152059049+ 512059049+ 102459049) = 5905959049 = 1

As the fractions sum up to 1, the values are true.

From the above values, a histogram is created using Microsoft Excel and then exported to Microsoft Word:

According to the histogram above, the modal value is 7 due to its highest probability. It would not be likely for Adam to get more than 8 points and less than 4 points by looking at the histogram above. Now using the binomial distribution, we can calculate the expected value and the standard distribution: Expected value = E (x) = np

= 10 ×23

≈6.667

Variance = σ2 = np (1-p)

= 203 (1 - 23) = 209

Therefore, Standard Deviation = σ

= 209 ≈1.4907

It is necessary to keep in mind the assumptions we have made over here. In the real game, there are many factors that influence the winning of an individual in the game, such as confidence, injury, stamina, etc., which would affect the game, and so player’s point probability doesn’t remain even. With these stipulations in mind, however, we...

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